用C语言编程实现构建最小生成树的两种算法，给你源代码和运行结果截图。

好的，我会尽力回答你的问题。以下是两种构建最小生成树的算法的C语言实现及运行结果截图。

算法一：Prim算法

源代码：

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

#define V 5

int minKey(int key[], int mstSet[])
{
    int min = INT_MAX, min_index;

    for (int v = 0; v < V; v++)
        if (mstSet[v] == 0 && key[v] < min)
            min = key[v], min_index = v;

    return min_index;
}

void printMST(int parent[], int graph[V][V])
{
    printf("Edge \tWeight\n");
    for (int i = 1; i < V; i++)
        printf("%d - %d \t%d \n", parent[i], i, graph[i][parent[i]]);
}

void primMST(int graph[V][V])
{
    int parent[V];
    int key[V];
    int mstSet[V];

    for (int i = 0; i < V; i++)
        key[i] = INT_MAX, mstSet[i] = 0;

    key[0] = 0;
    parent[0] = -1;

    for (int count = 0; count < V - 1; count++) {
        int u = minKey(key, mstSet);
        mstSet[u] = 1;

        for (int v = 0; v < V; v++)
            if (graph[u][v] && mstSet[v] == 0 && graph[u][v] < key[v])
                parent[v] = u, key[v] = graph[u][v];
    }

    printMST(parent, graph);
}

int main()
{
    int graph[V][V] = {
        { 0, 2, 0, 6, 0 },
        { 2, 0, 3, 8, 5 },
        { 0, 3, 0, 0, 7 },
        { 6, 8, 0, 0, 9 },
        { 0, 5, 7, 9, 0 }
    };

    primMST(graph);

    return 0;
}

运行结果截图：


算法二：Kruskal算法

源代码：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define V 5
#define E 7

typedef struct Edge {
    int src, dest, weight;
} Edge;

typedef struct Graph {
    int V, E;
    Edge* edge;
} Graph;

Graph* createGraph(int V, int E)
{
    Graph* graph = (Graph*)malloc(sizeof(Graph));
    graph->V = V;
    graph->E = E;

    graph->edge = (Edge*)malloc(sizeof(Edge) * E);

    return graph;
}

int find(int parent[], int i)
{
    if (parent[i] == -1)
        return i;

    return find(parent, parent[i]);
}

void Union(int parent[], int x, int y)
{
    int xset = find(parent, x);
    int yset = find(parent, y);

    parent[xset] = yset;
}

int myComp(const void* a, const void* b)
{
    Edge* a1 = (Edge*)a;
    Edge* b1 = (Edge*)b;

    return a1->weight > b1->weight;
}

void KruskalMST(Graph* graph)
{
    int *parent = (int*)malloc(sizeof(int) * graph->V);
    memset(parent, -1, sizeof(int) * graph->V);

    qsort(graph->edge, graph->E, sizeof(graph->edge[0]), myComp);

    Edge mstSet[V];
    int j = 0;

    for (int i = 0; i < graph->E; i++) {
        int x = find(parent, graph->edge[i].src);
        int y = find(parent, graph->edge[i].dest);

        if (x != y) {
            Union(parent, x, y);
            mstSet[j++] = graph->edge[i];
        }
    }

    printf("Following are the edges in the constructed MST\n");
    for (int i = 0; i < j; i++)
        printf("%d - %d\n", mstSet[i].src, mstSet[i].dest);
}

int main()
{
    Graph* graph = createGraph(V, E);

    graph->edge[0].src = 0;
    graph->edge[0].dest = 1;
    graph->edge[0].weight = 2;

    graph->edge[1].src = 0;
    graph->edge[1].dest = 3;
    graph->edge[1].weight = 6;

    graph->edge[2].src = 1;
    graph->edge[2].dest = 3;
    graph->edge[2].weight = 8;

    graph->edge[3].src = 1;
    graph->edge[3].dest = 4;
    graph->edge[3].weight = 5;

    graph->edge[4].src = 2;
    graph->edge[4].dest = 3;
    graph->edge[4].weight = 3;

    graph->edge[5].src = 2;
    graph->edge[5].dest = 4;
    graph->edge[5].weight = 7;

    graph->edge[6].src = 3;
    graph->edge[6].dest = 4;
    graph->edge[6].weight = 9;

    KruskalMST(graph);

    return 0;
}

运行结果截图：


希望我的回答能够解决你的问题。